第 24226 題：實指數函數

原題與已知條件

E(x)=sum_n>=0 x^n/n!的完整coefficient law與exact rational certificate。操作=exp-finite-partial-sum；資料={"x":[1,2],"cutoff":5}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。

• G001 — E(x)=sum_n>=0 x^n/n!的完整coefficient law與exact rational certificate。操作=exp-finite-partial-sum；資料={"x":[1,2],"cutoff":5}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。

定義與適用範圍

沿用原題明示的論域、假設與來源 MKO 定義。本題已完成第二對話數理與中英語義覆核，並由原始參數逐題獨立重算模型、答案與每個算術步驟；原題仍為 candidate。

• mko-real-exponential-function

逐步推導

source roles = [exp-finite-partial-sum]

derived = [[x, cutoff, terms, partial_sums, finite_sum, infinite_value_certificate, boundary]]

source roles = [0]

derived = [1]

(1/2)^0

Exp ORIGINAL x power0 = 1

Exp term0 power over factorial := div(1, 1)

Exp term0 power over factorial = 1

Exp cumulative sum through0 := sum(0, 1)

Exp cumulative sum through0 = 1

Exp term1 factorial step1 := mul(1, 1)

Exp term1 factorial step1 = 1

(1/2)^1

Exp ORIGINAL x power1 = 1/2

Exp term1 power over factorial := div(1/2, 1)

Exp term1 power over factorial = 1/2

Exp cumulative sum through1 := sum(1, 1/2)

Exp cumulative sum through1 = 3/2

Exp term2 factorial step1 := mul(1, 1)

Exp term2 factorial step1 = 1

Exp term2 factorial step2 := mul(1, 2)

Exp term2 factorial step2 = 2

(1/2)^2

Exp ORIGINAL x power2 = 1/4

Exp term2 power over factorial := div(1/4, 2)

Exp term2 power over factorial = 1/8

Exp cumulative sum through2 := sum(3/2, 1/8)

Exp cumulative sum through2 = 13/8

Exp term3 factorial step1 := mul(1, 1)

Exp term3 factorial step1 = 1

Exp term3 factorial step2 := mul(1, 2)

Exp term3 factorial step2 = 2

Exp term3 factorial step3 := mul(2, 3)

Exp term3 factorial step3 = 6

(1/2)^3

Exp ORIGINAL x power3 = 1/8

Exp term3 power over factorial := div(1/8, 6)

Exp term3 power over factorial = 1/48

Exp cumulative sum through3 := sum(13/8, 1/48)

Exp cumulative sum through3 = 79/48

Exp term4 factorial step1 := mul(1, 1)

Exp term4 factorial step1 = 1

Exp term4 factorial step2 := mul(1, 2)

Exp term4 factorial step2 = 2

Exp term4 factorial step3 := mul(2, 3)

Exp term4 factorial step3 = 6

Exp term4 factorial step4 := mul(6, 4)

Exp term4 factorial step4 = 24

(1/2)^4

Exp ORIGINAL x power4 = 1/16

Exp term4 power over factorial := div(1/16, 24)

Exp term4 power over factorial = 1/384

Exp cumulative sum through4 := sum(79/48, 1/384)

Exp cumulative sum through4 = 211/128

Exp term5 factorial step1 := mul(1, 1)

Exp term5 factorial step1 = 1

Exp term5 factorial step2 := mul(1, 2)

Exp term5 factorial step2 = 2

Exp term5 factorial step3 := mul(2, 3)

Exp term5 factorial step3 = 6

Exp term5 factorial step4 := mul(6, 4)

Exp term5 factorial step4 = 24

Exp term5 factorial step5 := mul(24, 5)

Exp term5 factorial step5 = 120

(1/2)^5

Exp ORIGINAL x power5 = 1/32

Exp term5 power over factorial := div(1/32, 120)

Exp term5 power over factorial = 1/3840

Exp cumulative sum through5 := sum(211/128, 1/3840)

Exp cumulative sum through5 = 6331/3840

source roles = [1/2, 5, [1, 1/2, 1/8, 1/48, 1/384, 1/3840], [1, 3/2, 13/8, 79/48, 211/128, 6331/3840]]

derived = [6331/3840, 否, finite sum is not a general infinite-value certificate]

逐步反向核對

• S036 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S035 · premise_and_result_audit — 以精確有理數重算 Exp cumulative sum through5，不以小數近似替代。
• S034 · premise_and_result_audit — 以精確有理數重算 Exp term5 power over factorial，不以小數近似替代。
• S033 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S032 · premise_and_result_audit — 以精確有理數重算 Exp term5 factorial step5，不以小數近似替代。
• S031 · premise_and_result_audit — 以精確有理數重算 Exp term5 factorial step4，不以小數近似替代。
• S030 · premise_and_result_audit — 以精確有理數重算 Exp term5 factorial step3，不以小數近似替代。
• S029 · premise_and_result_audit — 以精確有理數重算 Exp term5 factorial step2，不以小數近似替代。
• S028 · premise_and_result_audit — 以精確有理數重算 Exp term5 factorial step1，不以小數近似替代。
• S027 · premise_and_result_audit — 以精確有理數重算 Exp cumulative sum through4，不以小數近似替代。
• S026 · premise_and_result_audit — 以精確有理數重算 Exp term4 power over factorial，不以小數近似替代。
• S025 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S024 · premise_and_result_audit — 以精確有理數重算 Exp term4 factorial step4，不以小數近似替代。
• S023 · premise_and_result_audit — 以精確有理數重算 Exp term4 factorial step3，不以小數近似替代。
• S022 · premise_and_result_audit — 以精確有理數重算 Exp term4 factorial step2，不以小數近似替代。
• S021 · premise_and_result_audit — 以精確有理數重算 Exp term4 factorial step1，不以小數近似替代。
• S020 · premise_and_result_audit — 以精確有理數重算 Exp cumulative sum through3，不以小數近似替代。
• S019 · premise_and_result_audit — 以精確有理數重算 Exp term3 power over factorial，不以小數近似替代。
• S018 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S017 · premise_and_result_audit — 以精確有理數重算 Exp term3 factorial step3，不以小數近似替代。
• S016 · premise_and_result_audit — 以精確有理數重算 Exp term3 factorial step2，不以小數近似替代。
• S015 · premise_and_result_audit — 以精確有理數重算 Exp term3 factorial step1，不以小數近似替代。
• S014 · premise_and_result_audit — 以精確有理數重算 Exp cumulative sum through2，不以小數近似替代。
• S013 · premise_and_result_audit — 以精確有理數重算 Exp term2 power over factorial，不以小數近似替代。
• S012 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S011 · premise_and_result_audit — 以精確有理數重算 Exp term2 factorial step2，不以小數近似替代。
• S010 · premise_and_result_audit — 以精確有理數重算 Exp term2 factorial step1，不以小數近似替代。
• S009 · premise_and_result_audit — 以精確有理數重算 Exp cumulative sum through1，不以小數近似替代。
• S008 · premise_and_result_audit — 以精確有理數重算 Exp term1 power over factorial，不以小數近似替代。
• S007 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S006 · premise_and_result_audit — 以精確有理數重算 Exp term1 factorial step1，不以小數近似替代。
• S005 · premise_and_result_audit — 以精確有理數重算 Exp cumulative sum through0，不以小數近似替代。
• S004 · premise_and_result_audit — 以精確有理數重算 Exp term0 power over factorial，不以小數近似替代。
• S003 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S002 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S001 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。

結論

(1/2, 5, [1, 1/2, 1/8, 1/48, 1/384, 1/3840], [1, 3/2, 13/8, 79/48, 211/128, 6331/3840], 6331/3840, 否, finite_partial_sum_is_not_the_infinite_exponential_value)

依上列完整代值及條件核對得到原題答案；原題編號、格式與狀態保持不變。

本題覆核與來源綁定

覆核日期：2026-10-26

原題、答案、步驟表達式、數值 AST 與既有歷史保持原樣；原來的候選狀態屬於來源歷史，不覆寫為形式化證明。